WORST_CASE(Omega(1),?)

### Pre-processing the ITS problem ###

Initial linear ITS problem
   Start location: __init
      0: f1 -> f2 : arg1'=arg1P_1, arg2'=arg2P_1, [ arg1==arg1P_1 && 1==arg2P_1 ], cost: 1
      1: f2 -> f3 : arg1'=arg1P_2, arg2'=arg2P_2, [ 10000==arg1P_2 && arg2==arg2P_2 ], cost: 1
      4: f3 -> f4 : arg1'=arg1P_5, arg2'=arg2P_5, [ -arg2+arg1>=1 && arg1==arg1P_5 && arg2==arg2P_5 ], cost: 1
      6: f3 -> f7 : arg1'=arg1P_7, arg2'=arg2P_7, [ -arg2+arg1<1 && arg1==arg1P_7 && arg2==arg2P_7 ], cost: 1
      2: f4 -> f5 : arg1'=arg1P_3, arg2'=arg2P_3, [ arg2P_3==1+arg2 && arg1==arg1P_3 ], cost: 1
      3: f5 -> f6 : arg1'=arg1P_4, arg2'=arg2P_4, [ arg1P_4==-1+arg1 && arg2==arg2P_4 ], cost: 1
      5: f6 -> f3 : arg1'=arg1P_6, arg2'=arg2P_6, [ arg1==arg1P_6 && arg2==arg2P_6 ], cost: 1
      7: __init -> f1 : arg1'=arg1P_8, arg2'=arg2P_8, [], cost: 1

Checking for constant complexity:
   The following rule is satisfiable with cost >= 1, yielding constant complexity:
      7: __init -> f1 : arg1'=arg1P_8, arg2'=arg2P_8, [], cost: 1

Removed unreachable and leaf rules:
   Start location: __init
      0: f1 -> f2 : arg1'=arg1P_1, arg2'=arg2P_1, [ arg1==arg1P_1 && 1==arg2P_1 ], cost: 1
      1: f2 -> f3 : arg1'=arg1P_2, arg2'=arg2P_2, [ 10000==arg1P_2 && arg2==arg2P_2 ], cost: 1
      4: f3 -> f4 : arg1'=arg1P_5, arg2'=arg2P_5, [ -arg2+arg1>=1 && arg1==arg1P_5 && arg2==arg2P_5 ], cost: 1
      2: f4 -> f5 : arg1'=arg1P_3, arg2'=arg2P_3, [ arg2P_3==1+arg2 && arg1==arg1P_3 ], cost: 1
      3: f5 -> f6 : arg1'=arg1P_4, arg2'=arg2P_4, [ arg1P_4==-1+arg1 && arg2==arg2P_4 ], cost: 1
      5: f6 -> f3 : arg1'=arg1P_6, arg2'=arg2P_6, [ arg1==arg1P_6 && arg2==arg2P_6 ], cost: 1
      7: __init -> f1 : arg1'=arg1P_8, arg2'=arg2P_8, [], cost: 1

Simplified all rules, resulting in:
   Start location: __init
      0: f1 -> f2 : arg2'=1, [], cost: 1
      1: f2 -> f3 : arg1'=10000, [], cost: 1
      4: f3 -> f4 : [ -arg2+arg1>=1 ], cost: 1
      2: f4 -> f5 : arg2'=1+arg2, [], cost: 1
      3: f5 -> f6 : arg1'=-1+arg1, [], cost: 1
      5: f6 -> f3 : [], cost: 1
      7: __init -> f1 : arg1'=arg1P_8, arg2'=arg2P_8, [], cost: 1

### Simplification by acceleration and chaining ###

Eliminated locations (on linear paths):
   Start location: __init
     12: f3 -> f3 : arg1'=-1+arg1, arg2'=1+arg2, [ -arg2+arg1>=1 ], cost: 4
      9: __init -> f3 : arg1'=10000, arg2'=1, [], cost: 3

Accelerating simple loops of location 2.
   Accelerating the following rules:
     12: f3 -> f3 : arg1'=-1+arg1, arg2'=1+arg2, [ -arg2+arg1>=1 ], cost: 4

   Accelerated rule 12 with backward acceleration, yielding the new rule 13.
[accelerate] Nesting with 1 inner and 1 outer candidates
   Removing the simple loops: 12.

Accelerated all simple loops using metering functions (where possible):
   Start location: __init
     13: f3 -> f3 : arg1'=-k+arg1, arg2'=k+arg2, [ k>=0 && 2-2*k-arg2+arg1>=1 ], cost: 4*k
      9: __init -> f3 : arg1'=10000, arg2'=1, [], cost: 3

Chained accelerated rules (with incoming rules):
   Start location: __init
      9: __init -> f3 : arg1'=10000, arg2'=1, [], cost: 3
     14: __init -> f3 : arg1'=10000-k, arg2'=1+k, [ k>=0 && 10001-2*k>=1 ], cost: 3+4*k

Removed unreachable locations (and leaf rules with constant cost):
   Start location: __init
     14: __init -> f3 : arg1'=10000-k, arg2'=1+k, [ k>=0 && 10001-2*k>=1 ], cost: 3+4*k

### Computing asymptotic complexity ###

Fully simplified ITS problem
   Start location: __init
     14: __init -> f3 : arg1'=10000-k, arg2'=1+k, [ k>=0 && 10001-2*k>=1 ], cost: 3+4*k

Computing asymptotic complexity for rule 14
   Resulting cost 0 has complexity: Unknown

Obtained the following overall complexity (w.r.t. the length of the input n):
   Complexity:  Constant
   Cpx degree:  0
   Solved cost: 1
   Rule cost:   1
   Rule guard:  []

WORST_CASE(Omega(1),?)